\(\int \frac {(a+b x^2)^2}{x^7 \sqrt {c+d x^2}} \, dx\) [647]

Optimal result

Integrand size = 24, antiderivative size = 151 \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}-\frac {\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^3 x^2}+\frac {d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{7/2}} \]

[Out]

1/16*d*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(7/2)-1/6*a^2*(d*x^2+c)^(1/2)/c/x^6
-1/24*a*(-5*a*d+12*b*c)*(d*x^2+c)^(1/2)/c^2/x^4-1/16*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*(d*x^2+c)^(1/2)/c^3/x^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 91, 79, 44, 65, 214} \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=\frac {d \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{7/2}}-\frac {\sqrt {c+d x^2} \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right )}{16 c^3 x^2}-\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a \sqrt {c+d x^2} (12 b c-5 a d)}{24 c^2 x^4} \]

[In]

Int[(a + b*x^2)^2/(x^7*Sqrt[c + d*x^2]),x]

[Out]

-1/6*(a^2*Sqrt[c + d*x^2])/(c*x^6) - (a*(12*b*c - 5*a*d)*Sqrt[c + d*x^2])/(24*c^2*x^4) - ((8*b^2*c^2 - 12*a*b*
c*d + 5*a^2*d^2)*Sqrt[c + d*x^2])/(16*c^3*x^2) + (d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTanh[Sqrt[c + d*x^
2]/Sqrt[c]])/(16*c^(7/2))

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^2}{x^4 \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = -\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (12 b c-5 a d)+3 b^2 c x}{x^3 \sqrt {c+d x}} \, dx,x,x^2\right )}{6 c} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}+\frac {1}{16} \left (8 b^2-\frac {a d (12 b c-5 a d)}{c^2}\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = -\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}-\frac {\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^3 x^2}+\frac {\left (d \left (-8 b^2+\frac {a d (12 b c-5 a d)}{c^2}\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{32 c} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}-\frac {\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^3 x^2}+\frac {\left (-8 b^2+\frac {a d (12 b c-5 a d)}{c^2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{16 c} \\ & = -\frac {a^2 \sqrt {c+d x^2}}{6 c x^6}-\frac {a (12 b c-5 a d) \sqrt {c+d x^2}}{24 c^2 x^4}-\frac {\left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \sqrt {c+d x^2}}{16 c^3 x^2}+\frac {d \left (8 b^2-\frac {a d (12 b c-5 a d)}{c^2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=-\frac {\sqrt {c+d x^2} \left (24 b^2 c^2 x^4+12 a b c x^2 \left (2 c-3 d x^2\right )+a^2 \left (8 c^2-10 c d x^2+15 d^2 x^4\right )\right )}{48 c^3 x^6}+\frac {d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 c^{7/2}} \]

[In]

Integrate[(a + b*x^2)^2/(x^7*Sqrt[c + d*x^2]),x]

[Out]

-1/48*(Sqrt[c + d*x^2]*(24*b^2*c^2*x^4 + 12*a*b*c*x^2*(2*c - 3*d*x^2) + a^2*(8*c^2 - 10*c*d*x^2 + 15*d^2*x^4))
)/(c^3*x^6) + (d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(16*c^(7/2))

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(-\frac {-\frac {15 \left (a^{2} d^{2}-\frac {12}{5} a b c d +\frac {8}{5} b^{2} c^{2}\right ) x^{6} d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )}{8}+\left (\left (3 b^{2} x^{4}+3 a b \,x^{2}+a^{2}\right ) c^{\frac {5}{2}}+\frac {15 x^{2} \left (\left (-\frac {12 b \,x^{2}}{5}-\frac {2 a}{3}\right ) c^{\frac {3}{2}}+a \sqrt {c}\, d \,x^{2}\right ) d a}{8}\right ) \sqrt {d \,x^{2}+c}}{6 c^{\frac {7}{2}} x^{6}}\)	\(117\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (15 a^{2} d^{2} x^{4}-36 x^{4} a b c d +24 b^{2} c^{2} x^{4}-10 a^{2} c d \,x^{2}+24 a b \,c^{2} x^{2}+8 a^{2} c^{2}\right )}{48 c^{3} x^{6}}+\frac {\left (5 a^{2} d^{2}-12 a b c d +8 b^{2} c^{2}\right ) d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{16 c^{\frac {7}{2}}}\)	\(131\)
default	\(a^{2} \left (-\frac {\sqrt {d \,x^{2}+c}}{6 c \,x^{6}}-\frac {5 d \left (-\frac {\sqrt {d \,x^{2}+c}}{4 c \,x^{4}}-\frac {3 d \left (-\frac {\sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )+b^{2} \left (-\frac {\sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 c^{\frac {3}{2}}}\right )+2 a b \left (-\frac {\sqrt {d \,x^{2}+c}}{4 c \,x^{4}}-\frac {3 d \left (-\frac {\sqrt {d \,x^{2}+c}}{2 c \,x^{2}}+\frac {d \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )\)	\(227\)

[In]

int((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/c^(7/2)*(-15/8*(a^2*d^2-12/5*a*b*c*d+8/5*b^2*c^2)*x^6*d*arctanh((d*x^2+c)^(1/2)/c^(1/2))+((3*b^2*x^4+3*a*
b*x^2+a^2)*c^(5/2)+15/8*x^2*((-12/5*b*x^2-2/3*a)*c^(3/2)+a*c^(1/2)*d*x^2)*d*a)*(d*x^2+c)^(1/2))/x^6

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=\left [\frac {3 \, {\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \sqrt {c} x^{6} \log \left (-\frac {d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} - 12 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, c^{4} x^{6}}, -\frac {3 \, {\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \sqrt {-c} x^{6} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, a^{2} c^{3} + 3 \, {\left (8 \, b^{2} c^{3} - 12 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} x^{4} + 2 \, {\left (12 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c^{4} x^{6}}\right ] \]

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*sqrt(c)*x^6*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x
^2) - 2*(8*a^2*c^3 + 3*(8*b^2*c^3 - 12*a*b*c^2*d + 5*a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d
*x^2 + c))/(c^4*x^6), -1/48*(3*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt(-c)/sqrt(d*x^
2 + c)) + (8*a^2*c^3 + 3*(8*b^2*c^3 - 12*a*b*c^2*d + 5*a^2*c*d^2)*x^4 + 2*(12*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt
(d*x^2 + c))/(c^4*x^6)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (146) = 292\).

Time = 50.93 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.99 \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=- \frac {a^{2}}{6 \sqrt {d} x^{7} \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a^{2} \sqrt {d}}{24 c x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {5 a^{2} d^{\frac {3}{2}}}{48 c^{2} x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {5 a^{2} d^{\frac {5}{2}}}{16 c^{3} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {5 a^{2} d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{16 c^{\frac {7}{2}}} - \frac {a b}{2 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {a b \sqrt {d}}{4 c x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {3 a b d^{\frac {3}{2}}}{4 c^{2} x \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a b d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{4 c^{\frac {5}{2}}} - \frac {b^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 c x} + \frac {b^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2 c^{\frac {3}{2}}} \]

[In]

integrate((b*x**2+a)**2/x**7/(d*x**2+c)**(1/2),x)

[Out]

-a**2/(6*sqrt(d)*x**7*sqrt(c/(d*x**2) + 1)) + a**2*sqrt(d)/(24*c*x**5*sqrt(c/(d*x**2) + 1)) - 5*a**2*d**(3/2)/
(48*c**2*x**3*sqrt(c/(d*x**2) + 1)) - 5*a**2*d**(5/2)/(16*c**3*x*sqrt(c/(d*x**2) + 1)) + 5*a**2*d**3*asinh(sqr
t(c)/(sqrt(d)*x))/(16*c**(7/2)) - a*b/(2*sqrt(d)*x**5*sqrt(c/(d*x**2) + 1)) + a*b*sqrt(d)/(4*c*x**3*sqrt(c/(d*
x**2) + 1)) + 3*a*b*d**(3/2)/(4*c**2*x*sqrt(c/(d*x**2) + 1)) - 3*a*b*d**2*asinh(sqrt(c)/(sqrt(d)*x))/(4*c**(5/
2)) - b**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(2*c*x) + b**2*d*asinh(sqrt(c)/(sqrt(d)*x))/(2*c**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=\frac {b^{2} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{2 \, c^{\frac {3}{2}}} - \frac {3 \, a b d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{4 \, c^{\frac {5}{2}}} + \frac {5 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{16 \, c^{\frac {7}{2}}} - \frac {\sqrt {d x^{2} + c} b^{2}}{2 \, c x^{2}} + \frac {3 \, \sqrt {d x^{2} + c} a b d}{4 \, c^{2} x^{2}} - \frac {5 \, \sqrt {d x^{2} + c} a^{2} d^{2}}{16 \, c^{3} x^{2}} - \frac {\sqrt {d x^{2} + c} a b}{2 \, c x^{4}} + \frac {5 \, \sqrt {d x^{2} + c} a^{2} d}{24 \, c^{2} x^{4}} - \frac {\sqrt {d x^{2} + c} a^{2}}{6 \, c x^{6}} \]

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

1/2*b^2*d*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(3/2) - 3/4*a*b*d^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(5/2) + 5/16*a^2
*d^3*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(7/2) - 1/2*sqrt(d*x^2 + c)*b^2/(c*x^2) + 3/4*sqrt(d*x^2 + c)*a*b*d/(c^2*
x^2) - 5/16*sqrt(d*x^2 + c)*a^2*d^2/(c^3*x^2) - 1/2*sqrt(d*x^2 + c)*a*b/(c*x^4) + 5/24*sqrt(d*x^2 + c)*a^2*d/(
c^2*x^4) - 1/6*sqrt(d*x^2 + c)*a^2/(c*x^6)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=-\frac {\frac {3 \, {\left (8 \, b^{2} c^{2} d^{2} - 12 \, a b c d^{3} + 5 \, a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{3}} + \frac {24 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} d^{2} - 48 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} d^{2} + 24 \, \sqrt {d x^{2} + c} b^{2} c^{4} d^{2} - 36 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d^{3} + 96 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} d^{3} - 60 \, \sqrt {d x^{2} + c} a b c^{3} d^{3} + 15 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{4} - 40 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{4} + 33 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{4}}{c^{3} d^{3} x^{6}}}{48 \, d} \]

[In]

integrate((b*x^2+a)^2/x^7/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/48*(3*(8*b^2*c^2*d^2 - 12*a*b*c*d^3 + 5*a^2*d^4)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/(sqrt(-c)*c^3) + (24*(d*x
^2 + c)^(5/2)*b^2*c^2*d^2 - 48*(d*x^2 + c)^(3/2)*b^2*c^3*d^2 + 24*sqrt(d*x^2 + c)*b^2*c^4*d^2 - 36*(d*x^2 + c)
^(5/2)*a*b*c*d^3 + 96*(d*x^2 + c)^(3/2)*a*b*c^2*d^3 - 60*sqrt(d*x^2 + c)*a*b*c^3*d^3 + 15*(d*x^2 + c)^(5/2)*a^
2*d^4 - 40*(d*x^2 + c)^(3/2)*a^2*c*d^4 + 33*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(c^3*d^3*x^6))/d

Mupad [B] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+b x^2\right )^2}{x^7 \sqrt {c+d x^2}} \, dx=\frac {\frac {{\left (d\,x^2+c\right )}^{5/2}\,\left (5\,a^2\,d^3-12\,a\,b\,c\,d^2+8\,b^2\,c^2\,d\right )}{16\,c^3}-\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (5\,a^2\,d^3-12\,a\,b\,c\,d^2+6\,b^2\,c^2\,d\right )}{6\,c^2}+\frac {\sqrt {d\,x^2+c}\,\left (11\,a^2\,d^3-20\,a\,b\,c\,d^2+8\,b^2\,c^2\,d\right )}{16\,c}}{3\,c\,{\left (d\,x^2+c\right )}^2-3\,c^2\,\left (d\,x^2+c\right )-{\left (d\,x^2+c\right )}^3+c^3}+\frac {d\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (5\,a^2\,d^2-12\,a\,b\,c\,d+8\,b^2\,c^2\right )}{16\,c^{7/2}} \]

[In]

int((a + b*x^2)^2/(x^7*(c + d*x^2)^(1/2)),x)

[Out]

(((c + d*x^2)^(5/2)*(5*a^2*d^3 + 8*b^2*c^2*d - 12*a*b*c*d^2))/(16*c^3) - ((c + d*x^2)^(3/2)*(5*a^2*d^3 + 6*b^2
*c^2*d - 12*a*b*c*d^2))/(6*c^2) + ((c + d*x^2)^(1/2)*(11*a^2*d^3 + 8*b^2*c^2*d - 20*a*b*c*d^2))/(16*c))/(3*c*(
c + d*x^2)^2 - 3*c^2*(c + d*x^2) - (c + d*x^2)^3 + c^3) + (d*atanh((c + d*x^2)^(1/2)/c^(1/2))*(5*a^2*d^2 + 8*b
^2*c^2 - 12*a*b*c*d))/(16*c^(7/2))